Find the standard equation of a circle that satisfies the fo

Find the standard equation of a circle that satisfies the following condition: Endpoints of a diameter (-3, 2) and (5, -4).

Solution

The two end points of the diameter of the circle are (-3,2) and (5,-4) So the center of teh circle will be the mid point of the diameteric ends which is x_c= -3+5 / 2 = 2/2 = 1 and y_c= 2-4 / 2 = -2/2= -1;
Now the center of the circle is (1,-1) and diameter of the circle is distance between two end points of the diameter which is = sq rt [ (5-(-3) ) ² + (-4-2) ² ] = sq rt [ 8²+ (-6) ²] = sq rt ( 64+36) = sq rt (100) = 10;
Thus diameter = 10 means radius = 10/2 = 5
Hence the equation of the circle in standard form is (x-x_c)² + ( y-y_c)² = r2
Hence the equation of this circle is (x-1)²+ (y- (-1))²= 5² which is (x-1)²+ (y+1)²= 25